RULE 111.

For a Segment whofe Bafe is Perpendicular to the Fixed Axe; having given the Height, the Diameter of its Bafe, and a Diameter in the Middle between its Bafe and Vertex.

To the fquare of the diameter of the base, ald 4 times the fquare of the diameter in the middle, or the fquare of twice this diameter; multiply the fun by the height, and the product again by 13089969 for the content.

That is, (D2 + 4d2) x nb fegment; D being the diameter diameter in the middle, b the

*785398.*

Ꮓ

the content of the of the bafe, d the height, and n

E X

values being substituted in the above value of the fegment, gives Znh × (DD + 4dd) for the value of the fegment in terms of D, d, and h.

and

4dd-2DD

; which

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SCHOLIU M.

This theorem I have investigated in order to exprefs the value of this fegment independent of the axes of the fpheroid. For this purpose I was under a neceffity of introducing another dimenfion of the fegment, because it is not determinable from its bafe and height alone, as the other fegment was.

1

EXAMPLE 1.

What is the content of the fegment of an oblong fpheroid, whofe bafe is perpendicular to the fixed axe; its height being 5, the diameter of its bafe 18, and its middle diameter 319?

Here (DD+4dd)×4nb=[182 +(6√19)2] × 3 × *785398 (32+19) X 30 X 785398 210 X 314159659'73445 the content required.

EXAMPLE II.

Required the content of the fpheroidal fegment whofe height is 6, its bafe diameter 40, and diameter in the middle 30; the bafe being perpendicular to the fixed axe of the fpheroid.

Here (DD+4dd) × 1hp = (402+602)×X785398 = (22 + 33) × 202 X •785398 = 1300 X 3*14159 4084 07044 the content required.

PROBLEM. XV.

To find the Content of the Second Segment of a Spheroid.

As a fphere is to the fpheroid, fo is any part of the fphere to the like part of the fpheroid.*

PRO

* DEMONSTRATION.

For the like parts of any quantities are as the wholes. And that the fecond fegments, FGH, fgh, are like parts of the sphere and fpheroid, is evident from the nature of the figures.

Corol.

PROBLEM XVI.

To find the Surface of an Elliptic Spindle; or of any Fruftum or Segment of it.

Corol. 1. And if the fphere AKBI and fpheroid Akвi have one axe AB common; then the whole folids and the fecond fegments FGH, fgh, will be to each other, as the fquares of the other axes KI, ki, if the common axe AB be the fixed one. Or they will be to each other, as the other axes KI, ki, fimply, if the common axe be the revolving one. For, if a be the axe of the fphere, F and R the fixed and revolving axes of the fpheroid; the bodies will be to each other as A3 to FR2; hence if A = F, they will be as A2 to R2; but if A = R, they will be as A to F.

Corol. 2. Hence may be found the true quantity of liquor in a fpheroidal cafk, not full, whofe axe is parallel to the horizon. For if from the fegment akf be taken the double of the fecond fegment fgh, there will remain the part bhge; which taken from the whole cafk bbed will leave the part cdeg.

And after the fame manner may be found the quantity of liquor in a fpheroidal cafk partly filled, and ftanding a-tilt, with its axe inclined to the horizon.

the furface of the part of the fpindle generated by 'the arc BK about ad parallel to AD the fixed axe of the fpheroid.

EXAMPLE 1.

Given the axes of an ellipfe 50 and 40, to find the furface of the fpindle generated from an arc of that ellipfe, the length of the fpindle being 30.

50, BP = 40, ad eb = 30, Ae =

25 15 = 10, and eD = CD + cb =

*

DEMONSTRATION.

25

For the fluxions of the spindular furface is = the fluxion of the are BK drawn into iK × 2p = 2p% × (IK −1¿); and the fluent is equal to the fpheroidal furface BKMP —

or CH.

Corol. 1. The fpindular furface generated by EK, is equal to the fpheroidal furface BKMP the furface of a sphere whofe axe is a mean proportional between вк and 2CH. For 2pzX CH that fpheric surface.

Or 2px × CH = a circle whofe radius is

2CH × bk.

Corol. 2. When CH is equal to nothing, the spindle becomes barely a fpheroid: And when н falls below c, the furface of the fphere must be added.

What has been hitherto done, anfwers to the furface generated by an are about a line parallel to either axe of the ellipfe.

Corol. 3. From B the extremity of the lefs axe of the ellipfe, apply to AC the femi-tranfverfe; and parallel thereto draw AR meeting CB, produced in R; with the center c and radius AR defcribe the arc EF meeting the parallels aE, CB, 1к, bd in F, N, L, F: Draw CL and CF, and perpendicular to them iv and bs.

Then the circle whofe radius is a mean proportional between the fum and difference of two lines, of which the one is a mean proportional

25+15 = 40. Then as AC: CB::AeXеD: ea CH 16, the distance of the centers of the ellipfe and spindle.

By example 1 to problem 10, the fpheroidal furface generated by the arc ad, is 3686.8197024; and by problem 4, the length of the elliptic arc ad is 30.852.

585-2325 is the furface

Therefore 3686.8197 - 30.852 × 31416 X 32 3686-8197-3101.5872 of the fpindle.

axe.

proportional between BC and IV + LN, and the other a mean proportional between BK and 2CH, will be equal to the fpindular furface generated by BK about ad parallel to the tranfverfe For the fpheroidal furface is equal to a circle whofe radius is a mean proportional between Bc and IV + LN, by rule 4 to prob. 9; and 2p × BK X CH is equal to a circle whofe radius is a mean proportional between BK and 2cH; but the fpindular furface is equal to the difference of thofe two quantities; and the difference of two circles is equal to a circle whofe radius is a mean proportional between the fum and difference of the radii of the two circles; therefore, &c.

And, in the fame manner, the furface of half the fpindle, generated by the arc вd, is equal to a circle whose radius is a mean proportional between the fum and difference of two lines, the one of which is a mean proportional between BC and bs + NF, and the other a mean proportional between B and 2Cн,

Corol. 4. In like manner, for the furface of the fpindle whofe axe is parallel to the lefs axe of the ellipfe: having conftructed the annexed figure, as in page 309, the fpindular furface generated by the arc Cм, about NQ, will be equal to a circle whofe radius is a mean proportional between the fum and difference of two lines, of which the one is a mean proportional between AC and HI, and the other a mean proportional between Cм and 2BQ.

This, like the laft corollary, will be evident by comparing what is done above, with what is done in page 309.

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